Question

Solve the equation
$\frac{3}{2}lo{g}_{1/4}(x+2{)}^2-3=lo{g}_{1/4}(4-x{)}^3+lo{g}_{1/4}(x+6{)}^3$

Answer 1

$\frac{3}{2}{\log }_{1/4}(x+2{)}^2-3={\log }_{1/4}(4-x{)}^3+{\log }_{1/4}(x+6{)}^3$

${\log }_{1/4}(x+2{)}^3-{\log }_{1/4}(1/4{)}^3={\log }_{1/4}(4-x{)}^3+{\log }_{1/4}(x+6{)}^3$

$⟹64(x+2{)}^3=(4-x{)}^3(x+6{)}^3$

$4(x+2)=(4-x)(x+6)⟹x^2+6x-16=0⟹x=-8,2$

$x=-8$ is not possible $(\because x+2>0)$

$\therefore x=2$